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19 tháng 1 2020

1a Để \(\frac{x+1}{2}\)=\(\frac{8}{x+1}\)

\(\Rightarrow\)x+1.(x+1)=2.8=16

\(\Rightarrow\)x+1(x+1)=4.4

suy ra x+1=4

x=4-1

x=3

18 tháng 2 2020

a)(x+1)(x+1)=16

(x+1)^2=4^2

+)x+1=4

x=3

+)x+1=-4

x=-5

25 tháng 2 2017

a) \(\left|2x-1\right|=5\)

\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\left[\begin{matrix}=3\\=-2\end{matrix}\right.\)

b) \(\left(5^x-1\right)3-2=70\)

\(\Rightarrow5^x.3-3=72\)

\(\Rightarrow5^x.3=75\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

c) \(\left(x-1\frac{1}{2}\right)^2+\frac{3}{4}=\frac{1}{4}\)

\(\Rightarrow\left(x-1\frac{1}{2}\right)^2=\frac{-1}{2}\)

............. Làm tiếp nhé!

d) \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\) \(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\) \(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{23}{45}\) \(\Rightarrow\frac{22}{45}x=\frac{23}{45}\) \(\Rightarrow x=\frac{23}{45}:\frac{22}{45}\) \(\Rightarrow x=\frac{23}{22}\) Vậy \(x=\frac{23}{22}.\)
25 tháng 2 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{23}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\)

\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}\)

\(\Rightarrow x=\frac{23}{11}\)

16 tháng 3 2017

\(pt\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}.\frac{22}{45}.x=\frac{22}{45}\)

\(\Leftrightarrow\frac{1}{2}x=1\)

\(\Rightarrow x=2\)

25 tháng 2 2016

=>(2/1.2.3+2/2.3.4+....+2/8.9.10).x=22/45

=>(1/1.2-1/2.3+1/2.3-1/3.4+....+1/8.9-1/9.10).x=22/45

=>(1/1.2-1/9.10).x=22/45

=>22/45.x=44/45

=>x=2

25 tháng 2 2016

4+2^2+2^3+....+2^20=2^n

=>2^2+2^2+2^3+....+2^20=2^n

đặt 2^2+2^3+....+2^20

=>2A-A=2^21-2^2

khi đó A=2^2+2^21-2^2=2^21=2^n

=>n=21

31 tháng 1 2016

Đặt B = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{8.9.10}\)

=> 2B = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{8.9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\)

=> 2B = \(\frac{1}{1.2}-\frac{1}{9.10}\)

2B = \(\frac{22}{45}\)

B = \(\frac{22}{45}:2\)

=> B = \(\frac{11}{45}\)

Ta có : \(\frac{11}{45}.x=\frac{22}{45}\)

=> x = \(\frac{22}{45}:\frac{11}{45}\)

=> x = \(\frac{2}{1}\)

14 tháng 5 2017

a, \(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=0+\frac{1}{16}\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2=\left(\frac{-1}{4}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{3}=\frac{-1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{11}{12}\\\frac{1}{x}=\frac{5}{12}\end{cases}\Rightarrow\orbr{\begin{cases}11x=12\\5x=12\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}}\)

b, \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

Đặt S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\)

2S = \(\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{8.9.10}\)

2S = \(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\)

2S = \(\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)

S = \(\frac{22}{45}:2=\frac{11}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}\Rightarrow x=\frac{23}{11}\)

14 tháng 5 2017

a/ (1/x -2/3)2=1/16=(1/4)2

Có 2 trường hợp:

+/ 1/x -2/3= - 1/4

<=> 1/x =2/3 -1/4 = 5/12

=> x1=12/5

+/ 1/x - 2/3 =1/4

<=> 1/x = 2/3 +1/4= 11/12

=> x2=12/11

b/ Ta có: 

2/(1.2.3)=1/(1.2) - 1/2.3 ;  2/(2.3.4)=1/2.3 -1/3.4 ; ...; 2/(8.9.10)=1/8.9 -1/9.10

=> (1/1.2.3 + 1/2.3.4 +...+1/8.9.10)=23/45

<=> (1/1.2 -1/2.3 +1/2.3 -1/3.4 +...+1/8.9-1/9.10).x/2=23/45

<=> (1/1.2 -1/9.10).x/2 =23/45

<=> x.11/45=23/45

=> x=23/11

19 tháng 2 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{22}{45}x=\frac{44}{45}\)

\(\Rightarrow x=\frac{44}{45}:\frac{22}{45}\)

\(\Rightarrow x=2\)

Vậy x = 2

19 tháng 2 2017

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right):2=\frac{22}{45}:x\)

\(\Rightarrow\frac{22}{45}:2=\frac{22}{45}:x\)

\(\Rightarrow x=2\)

5 tháng 12 2015

\(\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).x=\frac{44}{45}\)

\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{44}{45}\)

\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{44}{45}\)

\(\frac{22}{45}.x=\frac{44}{45}\)

x =2

 

5 tháng 12 2015

X = 2 

9 tháng 10 2020

e. \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}+\frac{1}{5}-\frac{1}{45}=\frac{29}{45}\)

\(\Rightarrow\frac{7}{x}=\frac{7}{15}\)

\(\Rightarrow x=15\)

f. \(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{22}{45}\)

\(\Rightarrow\frac{1}{2}.\frac{22}{45}x=\frac{22}{45}\)

\(\Rightarrow\frac{11}{45}x=\frac{22}{45}\)

\(\Rightarrow x=2\)